How do I cite this document and does it say what I think it says?
#11
(08/15/2018, 03:33 AM)Chenjesu Wrote: ....
But more importantly, what does the half-exponential function actually look like in terms of functions? Trappman suggests something that looks like a basic composite of logs and exponentials, like you just exponentiate an iterated logarithm summed with a constant, yet I have not found a simplification that proves f(x) for f(f(x))=e^x.

I found this post on the halfexp function, halfexp(halfexp(z))=exp(z)
https://math.eretrandre.org/tetrationfor...hp?tid=544
Also, in the fatou.gp pari-gp program, halfsexp(z) is defined for complex values of z.  Here is a graph of the real axis from -10 to +10.  halfexp(20)=398.247512, halfexp(100)=192708.572  I have no idea how to express halfexp(z) in terms of a "composite of logs and exponentials".
[Image: half_exp_e.png]
- Sheldon
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#12
So then what is the significance of Trappmann's formulas then if they don't let you write what the half exponential function actually does to a number? Was that entire paper simply to suggest that there is a relationship between fractional heights and fractional iterations which is already a common-sense conjecture anyway?
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#13
(08/15/2018, 04:36 AM)Chenjesu Wrote: So then what is the significance of Trappmann's formulas then if they don't let you write what the half exponential function actually does to a number? What that entire paper simply to suggest that there is a relationship between fractional heights and fractional iterations which is already a common-sense conjecture anyway?

Trapmann proves a uniqueness criterion for the inverse of tetration.  Given a tetration solution for a real base>exp(1/e), one can generate an alternative analytic solution which also meets the criterion TetAlt(z+1)=exp_b(TetAlt(z))
\( \text{TetAlt}(z)=\text{Tet}(z+\theta(z)) \)
where theta(z) is a 1-cyclic fourier series.  Trapmann proves than any such alternate solution will not meet his uniqueness criteria for the inverse of tetration, and will not behave as nicely in the complex plane as Kneser's solution.  Trapmann's proof works for the Abel function, or slog, or inverse of tetration.
- Sheldon
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#14
Okay, so then what is the solution to the half-exponential that allows it to be plotted in the complex plane? Or otherwise how does this tetration relationship allow it to be plotted?
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#15
(08/16/2018, 12:24 AM)Chenjesu Wrote: Okay, so then what is the solution to the half-exponential that allows it to be plotted in the complex plane? Or otherwise how does this tetration relationship allow it to be plotted?

Assuming you have sexp(z) and its inverse the slog(z), you use sexp(slog(z)+0.5).  Then it depends a lot on the specific computational technique, and whether your program generates an slog or an sexp.  The most mathematically rigorous published computational technique is probably William Paulson's recent paper.

Also you can take advantage of halfexp(exp(z)) = exp(halfexp(z))
- Sheldon
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#16
But that directly contradicts what you said before that we don't actually know what the half-exp really is in terms of functions. So how do they graph a plot function with no explicit representation? Do they use a taylor series? hypergeometric series? bessel series? Or what?
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#17
(08/16/2018, 08:54 AM)Chenjesu Wrote: But that directly contradicts what you said before that we don't actually know what the half-exp really is in terms of functions. So how do they graph a plot function with no explicit representation? Do they use a taylor series? hypergeometric series? bessel series? Or what?

Actually, what I wrote is:
Quote:Unfortunately, the Riemann mapping is hard to compute, so realistically, all that we have is the proof of the existence of a nicely behaved solution.  In general, Riemann mappings cannot be expressed in terms of elementary functions or series and therefore Kneser's solution probably isn't expressible in terms of elementary functions. 

..  Paulsen's paper gives numerical results just like Trapmann's paper ... There are a few really good programs that can easily calculate sexp for any base, like my pari-gp program, fatou.gp  

My program iterates, generating more and more accurate representations for two different functions; one function is a Fourier series involving the Schroder function which is mathematically equivalent to Kneser's Riemann mapping.  The 1-cyclic Fourier series is mapping this image (equivalent to Figure 1 from Paulson's paper) to the real axis.  The region above is mapped to the upper half of the complex plane.  The 1-cyclic Fourier series does not converges near the singularities at the real axis, so near the real axis one can use a 2nd function.  The 2nd function involved in the iteration sequence is an approximation for the Taylor series for the slog, centered at a point on the real axis.  The two functions combined allow one to generate the slog and sexp for any point in the complex plane.  If the two functions match each other (within the each function's error term) where both functions are well behaved, then as the error term goes to zero, the pair of functions converge to Kneser's solution.  In the limit, the 1-cyclic Fourier series becomes mathematically equivalent to Kneser's Riemann mapping.  
[Image: abel_real_axis.png]
- Sheldon
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#18
Okay, you keep mentioning over and over about this alleged method of computing such functions, but what actually are they?
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#19
(08/17/2018, 05:59 PM)Chenjesu Wrote: Okay, you keep mentioning over and over about this alleged method of computing such functions, but what actually are they?
I'm not sure what your question is.  Is your question how to load the program?  You would first need to download pari-gp; here is the link: https://pari.math.u-bordeaux.fr/download.html
And then my fatou.gp program is here: https://math.eretrandre.org/tetrationfor...p?tid=1017
download the program and run in it in pari-gp

Were you able to understand the descriptions in this post and the links I've posted?  Or are you asking for more specific detail?  It is important to thoroughly understand Wikipedia-Schroder's-equation as applied to the complex fixed point for "f(z)=e^z", since that is the basis upon which Kneser's Riemann mapping is built, and it is also the basis for my faou.gp program.  Then you have to understand Kneser's Riemann mapping and how it can be equivalently expressed as a 1-cyclic Fourier series that changes the complex valued Schroder function to a real valued slog.

The next paragraph is an expository summary, with no equations, to provide a brief high level overview.  Assuming you understand the background, then my fatou.gp program iterates a pair of functions, an slog Taylor series, and the 1-cyclic Fourier series theta mapping which modifies the Schroder function.   First the Schroder function is changed into the complex valued Abel function pictured in post#17.  In the picture below, my program's slog Taylor series approximation samples 60 equally spaced points around the middle circle.  The points in yellow are paired up with exp(z), one for one.  The points in brown are paired up with log(z), also one for one.  The points in pink (and green for the complex conjugate), are used to generate the 1-cyclic Fourier series theta mapping.  You can see the pink spiral towards the fixed point.  The slog has a singularity at the fixed point where it goes to imag(infinity), so the radius of convergence of the slog is the blue circle.  The pink points inside the smaller are used to generate a 1-cyclic mapping involving the Schroder equation, which is used to generate an accurate value for the pink sample points in the middle circle.  It turns out this can be either be approximately solved which is what I do, or used to generate a 60x60 simultaneous equation, which my program can also do.  These 60 sample points give more than double precision accurate results.  If you solve the simultaneous equation, then all of the points will match exactly with their yellow or brown counterparts.  The pink/green points will exactly match the 1-cyclic theta mapping.  To get more higher precision results, we keep adding more and more sample points to the circle, and to the theta mapping, to get whatever precision we would like for the slog, and for the theta mapping.

Since the Schroder equation modified by the 1-cyclic theta mapping is equivalent to Kneser's Riemann mapping, the end result is Kneser's slog calculated to whatever precision we would like. Inside the pink spiral, the Schroder equation modified by the 1-cyclic theta mapping is used to generate accurate results; inside the middle circle, the slog Taylor series gives accurate results.   Both functions give accurate results where the two representations overlap.  We can also iterate exp(z) or log(z) to get to a point represented accurately.  
[Image: base_e_theta60.png]

There's a lot more detail.  My hope is when I have enough time, I would eventually publish a proof that the systems of equations approach using equally spaced points for both the slog taylor series, and the 1-cyclic fourier series, can be shown to be exactly Kneser's slog in the limit as the sample points get arbitrarily close together.  This requires proving the system of equations for a finite number of sample points always has exactly one solution.  And it requires showing that the error term is predictable and can be rigorously bounded.

I would also highly recommend William Paulson's work.
William Paulson has an online calculator here: http://myweb.astate.edu/wpaulsen/tetcalc/tetcalc.html
William Paulson's paper's describe his algorithm: http://myweb.astate.edu/wpaulsen/tetration.html
- Sheldon
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#20
Perhaps, I can clear things up here a little on my end since one of the paper's in question is my own. 

First of all I apologize if this forum doesn't meet your standards, it is mostly dead now. Sheldon and I tend to be the few who actually post on here anymore. There isn't much action, so I rarely post here anymore, but occasionally something catches my eye.

First of all, my paper is numerically correct. If we want to define the hyper-operators for \( 1 < \alpha < e^{1/e} \) and \( \Re(z) > 0 \), we need look no further than the recursive definition

\( \alpha \uparrow^0 z = \alpha \cdot z \)

\( \Gamma(1-z)(\alpha \uparrow^{n+1} z) = \sum_{k=0}^\infty(\alpha \uparrow^{n}\alpha \uparrow^n...(k+1\,times)...\uparrow^n \alpha) \frac{(-1)^k}{k!(k+1-z)} + \int_{1}^\infty f_n(x)x^{-z}\,dx \)

where

\( f_n(x) = \sum_{k=0}^\infty\,(\alpha \uparrow^{n}\alpha \uparrow^n...(k+1\,times)...\uparrow^n \alpha) \frac{(-x)^k}{k!} \)


If you want to work with tetration, it's simple. Just set \( n=2 \) and there's your tetration, and a nifty formula for it. This I could prove in a heartbeat, just a coupla' lines of math. I extended this to arbitrary \( n \) and did so a little too swift and not without fault.

The paper that is available to the public does have gaps in its construction of the hyperoperators. I worked on remedying them for a year, and eventually did; sadly, I never got around to rewriting the paper or fixing the errors in complete form. I was a little discouraged by how little interest there was in the hyper operators, so I abandoned it to work on more palpable "mainstream" (if you will) problems.

The final claim of the  paper is in fact correct (I can prove it now, but it would take about 30 - 40 pages of reasoning I haven't gotten around to writing out clearly), it is only that I proved somethings within the paper with too much of a hand wave that forego some of the subtleties of the question. In short, I gave an erroneous statement which led to the result. But the erroneous statement can be tweaked slightly to to give a correct statement which still gives the result.

As a professional, I would not cite my paper. It was the first paper I ever wrote and I got ahead of myself very often. I've tried consistently to remove it from arXiv, but that's what arXiv does, it archives. I've been meaning to rewrite the paper, but I haven't found much time for it. Mainly because of how little interest journals expressed towards the problem, and that no one was likely to publish it (disregarding it had a structural error).

All and all, you're in the same boat I was in back when I started writing that paper. There were no real papers to cite. And the papers to cite, were either: obscure, hard to follow, narrow, or never really said what you wanted to be said. Although people will tell you, straightforwardly, and incredibly often "a solution to \( \text{sexp}(z) \) exists," but you'll be hard pressed to find three or more papers on the matter written in a professional manner. They were just in the hemisphere of knowledge, but weren't down to earth yet. And as you've complained, these papers tend to not be written fluidly with a broad appeal.

You're in uncharted waters here. This math is fringe math (as I like to call it). Even if there are papers, only a select few know about them. And then, there are a select few who know different papers from other select fews, as though there's a schism between the literature available for the few people working on the same problem. How things are being proved here is similar to how maps of North America were being charted in the late 17th century. They looked pretty close to North America, but weren't actually the truth--and everyone had a different looking map. Consider the qualms of defining the logarithm in Euler's day, no one really knew what was going on when they saw \( \log(-1) = \pi i \) but it also equals \( -\pi i \).

To conclude. Do not cite my paper. If you have questions regarding my paper and qualms or what have you, pm me. I'd be happy to answer any questions. Just, again, my paper isn't worth the paper it's written on (at least mathematically). When one brick is missing at the bottom of a tower, the entire thing collapses. This area of math is largely an oral thing at this point. It's the discussions which are making the field, not the actual written word. To prove this point, Sheldon and I are responsible for a proof that tetration with base \( 1 < \alpha < e^{1/e} \) is a completely monotone function. But neither of us bothered to write up a proof. Which is the lay of the land in these parts.

Regards, James.
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